Complexity 12 1 NonDeterministic Space Complexity Andrei Bulatov
Complexity 12 -1 Non-Deterministic Space Complexity Andrei Bulatov
Complexity 12 -2 Non-deterministic Machines Recall that if NT is a non-deterministic Turing Machine, then NT(x) denotes the tree of configurations which can be entered with input x, and NT accepts x if there is some accepting path in NT(x) Definition The space complexity of a non-deterministic Turing Machine NT is the function such that is the minimal number of cells visited in an accepting path of NT(x) if there is one, otherwise it is the minimal number of cells in the rejecting paths (If not all paths of NT(x) halt, then is undefined)
Complexity Nondeterministic Space Complexity Definition For any function f, we say that the nondeterministic space complexity of a decidable language L is in O(f) if there exists a nondeterministic Turing Machine NT which decides L, and constants and c such that for all inputs x with Definition The nondeterministic space complexity class NSPACE[f] is defined to be the class of all languages with nondeterministic space complexity in O(f) 12 -3
Complexity Definition of NPSPACE Definition 12 -4
Complexity Savitch’s Theorem Unlike time, it can easily be shown that nondeterminism does not reduce the space requirements very much: Theorem (Savitch) If s(n) log n, then Corollary PSPACE NPSPACE 12 -5
Complexity 12 -6 Proof (for s(n) n) • Let L be a language in NSPACE[s] • Let NT be a non-deterministic Turing Machine that decides L with space complexity s • Choose an encoding for the computation NT(x) that uses ks(|x|) symbols for each configuration • Let be the initial configuration, and configuration • Define a Boolean function reach(C, C , j) which is true if and on configuration C can be reached from configuration C in at mos steps • To decide whether or not x L we must determine whether or n is true be the accepting
Complexity We can calculate in space, using a divide-and-conquer algorithm: 1. If j=0 then if C=C', or C' can be reached from C in one step, then return true, else return false. 2. For each configuration C'', if reach(C, C'', j – 1) and reach(C'', C', j – 1), then return true. 3. Return false The depth of recursion is O(s(|x|)) and each recursive call requires O(s(|x|)) space for the parameters 12 -7
Complexity 12 -8 Logarithmic Space Since polynomial space is so powerful, it is natural to consider more restricted space complexity classes Even linear space is enough to solve Satisfiability Definition
Complexity 12 -9 Problems in L and NL What sort of problems are in L and NL? In logarithmic space we can store: • a fixed number of counters (up to length of input) • a fixed number of pointers to positions in the input string Therefore in deterministic log-space we can solve problems that require a fixed number of counters and/or pointers for solving; in non-deterministic log-space we can solve problems that require a fixed number of counters/pointers for verifying a solution
Complexity Examples (L) Palindromes: We need to keep two counters First count the number of 0 s, then count 1 s, subtracting from the previous number one by one. If the result is 0, accept; otherwise, reject. Brackets (if brackets in an expression positioned correctly): We need only a counter of brackets currently open. If this counter gets negative, reject; otherwise accept if and 12 -10
Complexity 12 -11 Examples (NL) The first problem defined on this course was Reachability¹ This can be solved by the following non-deterministic algorithm: • Define a counter and initialize it to the number of vertices in th • Define a pointer to hold the ``current vertex’’ and initialize it to the start vertex • While the counter is non-zero - If the current vertex equals the target vertex, return yes - Non-deterministically choose a vertex which is connected the current vertex Update the pointer to this vertex and decrement the counte - • Return no ¹Also known as Path
Complexity Reducing Problems We have seen that polynomial time reduction between problems is a very useful concept for studying relative complexity of problems. It allowed us to distinguish a class of problems, NP, which includes many important problems and is viewed as the class of hard problems We are going to do the same for space complexity classes: NL and PSPACE There is a problem: Polynomial time reduction is too powerful 12 -12
Complexity 12 -13 Log-Space Reduction A transducer is a 3 -tape Turing Machine such that • the first tape is an input tape, it is never overwritten • the second tape is a working tape • the third tape is an output tape, no instruction of the transiti function uses the content of this tape The space complexity of such a machine is the number of cells on the working tape visited during a computation A function is said to be log-space computable if there is a transducer computing f in O(log n)
Complexity Definition A language A is log-space reducible to a language B , denoted , if a log-space computable function f exists such that for all Note that a function computable in log-space is computable in polynomial time, so 12 -14
Complexity Completeness Definition A language L is said to be NL-complete if L NL and, for any A NL, Definition A language L is said to be P-complete if A P and, for any A P, 12 -15
Complexity NL-Completeness of REACHABITITY Theorem Reachability is NL-complete Corollary NL P Proof Idea For any non-deterministic log-space machine NT, and any input x, construct the graph NT(x). Its vertices are possible configurations of NT using at most log(|x|) cells on the working tape; its edges are possible transitions between configurations. Then NT accepts the input x if and only if the accepting configuration is reachable from the initial configuration 12 -16
Complexity 12 -17 Proof • Let A be a language in NL • Let NT be a non-deterministic Turing Machine that decides A with space complexity log n • Choose an encoding for the computation NT(x) that uses klog(|x|) symbols for each configuration • Let be the initial configuration, and configuration be the accepting • We represent NT(x) by giving first the list of vertices, and then a list of edges
Complexity • 12 -18 Our transducer T does the following - T goes through all possible strings of length klog(|x|) and, the string properly encodes a configuration of NT, prints it on the output tape - Then T goes through all possible pairs of strings of length klog(|x|). For each pair it checks if both strings a legal encodings of configurations of NT, and if can yie. If yes then it prints out the pair on the output tape • Both operations can be done in log-space because the first s requires storing only the current string (the strings can be liste in lexicographical order). Similarly, the second step requires storing two strings, and (possibly) some counters • NT accepts x if and only if there is a path in NT(x) from
Complexity 12 -19 Log-Space reductions and L We take it for granted that P is closed under polynomial-time reduc We can expect that L is closed under log-space reductions, but it is much less trivial Theorem If and B L, then A L Corollary If any NL-complete language belongs to L, then L = NL
Computability and Complexity 12 -20 Proof Let M be a Turing Machine solving B in log-space, and let T be a log-space transducer reducing A to B It is not possible to construct a log-space decider for A just combining M and T, because the output of T may require more than log-space Instead, we do the following Let f be the function computed by T On an input x, a decider M' for A • Simulates M on f(x) • When it needs to read the l-th symbol of f(x), M' simulates but ignores all outputs except for the l-th symbol
Complexity 12 -21 P-completeness Using log-space reductions we can study the finer structure of the c A clause is said to be a Horn if it contains at most one positive literal A CNF is said to be Horn if every its clause is Horn-SAT Instance: A Horn CNF . Question: Is satisfiable?
Complexity Theorem Horn-SAT is P-complete 12 -22
Complexity 12 -23 Time and Space All Languages Decidable Languages NP L NL NL-complete PSPACE P P-complete NP-complete
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